\(\left[\left(4x+28\right)\times3+55\right]\div5=35\)

\(\left(4x+28\right)\times3+55=175\)

\(12x+84+55=175\)

\(12x=175-55-84\)

\(12x=36\)

\(x=3\)

\(4x-\dfrac{5}{6}=2x+\dfrac{2}{3}\)

\(4x-\dfrac{5}{6}-2x-\dfrac{2}{3}=0\)

\(2x-\dfrac{3}{2}=0\)

\(2x=\dfrac{3}{2}\)

\(x=\dfrac{3}{4}\)

\(\left(\dfrac{2}{5}-\dfrac{1}{3}\right)^2\)

=  \(\left(\dfrac{1}{15}\right)^2\)

=  \(\dfrac{1}{225}\)

a) \(\dfrac{2}{9}+\dfrac{1}{5}+\dfrac{7}{9}+\dfrac{4}{5}\)

\(=\left(\dfrac{2}{9}+\dfrac{7}{9}\right)+\left(\dfrac{1}{5}+\dfrac{4}{5}\right)\)

\(=1+1\)

\(=2\)

b) Phân số chỉ số bóng vàng là:

\(1-\dfrac{1}{2}-\dfrac{1}{3}=\dfrac{1}{6}\)  ( số bóng vàng )

Nhiệt độ ở Ottawa lúc 10h là:

\(-4+6=2^oC\)

Nếu mà theo cách x - 50 = 0 thì bạn theo cách này nha:

\(\dfrac{x-17}{33}+\dfrac{x-21}{29}+\dfrac{x}{25}=4\)

\(\dfrac{x-17}{33}+\dfrac{x-21}{29}+\dfrac{x}{25}-4=0\)

\(\dfrac{x-17}{33}-1+\dfrac{x-21}{29}-1+\dfrac{x}{25}-2=0\)

\(\dfrac{x-50}{33}+\dfrac{x-50}{29}+\dfrac{x-50}{25}=0\)

\(\left(x-50\right)\left(\dfrac{1}{33}+\dfrac{1}{29}+\dfrac{1}{5}\right)=0\)

Vì  \(\dfrac{1}{33}+\dfrac{1}{29}+\dfrac{1}{25}>0\)

=> \(x-50=0\)
=> \(x=50\)

\(\dfrac{x-17}{33}+\dfrac{x-21}{29}+\dfrac{x}{25}=4\)

\(\dfrac{x-17}{33}+\dfrac{x-21}{29}+\dfrac{x}{25}-4=0\)

\(\dfrac{\left(x-17\right)\times725}{33\times725}+\dfrac{\left(x-21\right)\times825}{29\times825}+\dfrac{x\times957}{25\times957}-\dfrac{4\times23925}{23925}=0\)

\(725x-12325+825x-17325+957x-95700=0\)

\(2507x-125350=0\)

\(2507x=125350\)

\(x=50\)

\(\dfrac{y}{20}-\dfrac{7}{12}=\dfrac{11}{30}\)

\(\dfrac{y}{20}=\dfrac{11}{30}+\dfrac{7}{12}\)

\(\dfrac{y}{20}=\dfrac{19}{20}\)

\(y\times20=19\times20\)

\(y\times20=380\)

\(y=19\)

\(\left(\dfrac{1}{2}+1\right)\times\left(\dfrac{1}{3}+1\right)\times\left(\dfrac{1}{4}+1\right)\times\left(\dfrac{1}{5}+1\right)\times\left(\dfrac{1}{6}+1\right)\)

\(=\dfrac{3}{2}\times\dfrac{4}{3}\times\dfrac{5}{4}\times\dfrac{6}{5}\times\dfrac{7}{6}\)

\(=\dfrac{3\times4\times5\times6\times7}{2\times3\times4\times5\times6}\)

\(=\dfrac{7}{2}\)

\(34\times x+138=1137\)

\(34\times x=1137-138\)

\(34\times x=999\)

\(x=\dfrac{999}{34}\)

Vậy: \(x=\dfrac{999}{34}\)